Question: Simplify the following expression and state the condition under which the simplification is valid. $y = \dfrac{8q^3 + 112q^2 + 320q}{q^3 + q^2 - 90q}$
First factor out the greatest common factors in the numerator and in the denominator. $ y = \dfrac {8q(q^2 + 14q + 40)} {q(q^2 + q - 90)} $ $ y = \dfrac{8q}{q} \cdot \dfrac{q^2 + 14q + 40}{q^2 + q - 90} $ Simplify: $ y = 8 \cdot \dfrac{q^2 + 14q + 40}{q^2 + q - 90}$ Since we are dividing by $q$ , we must remember that $q \neq 0$ Next factor the numerator and denominator. $ y = 8 \cdot \dfrac{(q + 10)(q + 4)}{(q + 10)(q - 9)}$ Assuming $q \neq -10$ , we can cancel the $q + 10$ $ y = 8 \cdot \dfrac{q + 4}{q - 9}$ Therefore: $ y = \dfrac{ 8(q + 4)}{ q - 9 }$, $q \neq -10$, $q \neq 0$